Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4. What is the pH of this solution at equilibrium?

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

Let's solve this for x. Multiply both sides by 0.12

taking square root to both sides:

Now, we have got the concentration of

We know that,

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

The correct answer is 2.34

Explanation:

HCOOH is formic acid. It is a weak acid so it does not dissociates completely in water. At the beggining (I) the initial concentration is 0.12 M. In water it will dissociate in a certain grade x as follows:

          HCOOH    →    H⁺ + HCOO⁻

I             0.12 M           0          0                    

C           - x                   x          x

E          (0.12 M - x)       x          x

The mathematical expression for the equilibrium constant (Ka) is the following:

As the value of Ka is too small in comparison with the initial concentration 0.12 M, we can approximate: 0.12 M - X ≅ 0.12 M. Then, we calculate x:

1.8 x 10⁻⁴ = x²/0.12 M

⇒ x= = 4.65 x 10⁻³

Since x = 4.65 x 10⁻³ , from the equilibrium we have:

[H⁺] = x = 4.65 x 10⁻³

From the definition of pH, we have:

pH = -log [H⁺] = -log (4.65 x 10⁻³)= 2.34

there are no statements provided but here is why:

sulfur belongs to group 16 because it can be found in nature whether it is in a free or combined state.

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